Question

Determine the empirical formula of an oxide of iron
which has 69.9% iron and 30.1% dioxygen by mass
[Atomic mass of Fe = 56, 0=16]​

Answer 1

Answer:

empirical formula=Fe2O3

Explanation:

percentage of iron =69.9%

percentage of dioxygen=30.1%

now,

Mole of atom of iron (Fe) = 69.9 ÷ 56

=1.24

Mole of atom of oxygen(O) = 30.1 ÷ 16

= 1.88

again,

Mole ratio of iron = 1.24÷1.24 = 1

Mole ratio of oxygen=1.88÷1.24= 1.5

multiply both answers by 2 [since we need the smallest whole ratio] i,e. 1×2=2

1.5×2=3

the final answer is Fe2O3. (ion oxide).